Regents Algebra I Practice Questions (2026) — Free Sample Questions with Answers

Answer: C

1. 1. $xy=\sqrt{36}=6$ (rational).
2. 2. $\frac{x}{y}=\sqrt{9}=3$ (rational).
3. 3. $x+y=3\sqrt{2}+\sqrt{2}=4\sqrt{2}$ (irrational).
4. 4. $(x-y{)}^2=(2\sqrt{2}{)}^2=8$ (rational).

$xy$: $\sqrt{18}\cdot \sqrt{2}=\sqrt{36}=6$, which is rational.

$\frac{x}{y}$: $\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3$, which is rational.

$x+y$: Correct. $\sqrt{18}+\sqrt{2}=3\sqrt{2}+\sqrt{2}=4\sqrt{2}$, a nonzero rational number times an irrational number, so it is irrational.

$(x-y{)}^2$: $x-y=3\sqrt{2}-\sqrt{2}=2\sqrt{2}$, and $(2\sqrt{2}{)}^2=8$, which is rational.

Rewrite in simplest radical form first: a result with a leftover radical such as $4\sqrt{2}$ is irrational, while perfect squares under the radical give rational numbers.

Answer: A

1. Locate the intersection points: $(-2,5)$ and $(3,0)$.
2. Read the x-coordinates: $x=-2$ and $x=3$.

$-2$ and $3$: Correct. The graphs intersect at $(-2,5)$ and $(3,0)$; the x-coordinates of the intersection points are the solutions of $f(x)=g(x)$.

$-1$ and $3$: These are the zeros of $f$ (where $f(x)=0$), not the points where the two graphs meet.

$0$ and $5$: These are the y-coordinates of the intersection points; the question asks for $x$.

$3$ only: Uses only the intersection on the x-axis; the graphs also meet at $(-2,5)$.

The solutions of f(x) = g(x) are the x-coordinates of the points where the graphs of y = f(x) and y = g(x) intersect.

Answer: 7 weeks

1. 1. Inequality: $45+12w>153-6w$.
2. 2. Add $6w$ and subtract 45: $18w>108$.
3. 3. $w>6$.
4. 4. At $w=6$ they each have $117 (equal), so the first week Kendra has more is week 7: $129 versus $111.
5. Common errors: 6 (the week they are equal) or 10 (solves $45+12w>153$, ignoring Lia's spending).
6. On the Regents, a correct answer with no work shown earns only 1 of 2 credits.

"More than" is a strict inequality; the first whole number greater than the boundary value answers the question.

Answer: A

1. 1. $a=5$, $b=2$, $c=-1$.
2. 2. $b^2-4ac=4-4(5)(-1)=24$, and $\sqrt{24}=2\sqrt{6}$.
3. 3. $x=\frac{-2\pm 2\sqrt{6}}{2(5)}=\frac{-2\pm 2\sqrt{6}}{10}$.
4. 4. Divide numerator and denominator by 2: $\frac{-1\pm \sqrt{6}}{5}$.

$\frac{-1\pm \sqrt{6}}{5}$: Correct. $x=\frac{-2\pm \sqrt{4+20}}{10}=\frac{-2\pm 2\sqrt{6}}{10}=\frac{-1\pm \sqrt{6}}{5}$.

$\frac{1\pm \sqrt{6}}{5}$: Sign slip: $-b$ is $-2$, not $+2$.

$\frac{-1\pm \sqrt{6}}{10}$: Divides the numerator by 2 but not the denominator, so the fraction is not equivalent.

$\frac{-2\pm 2\sqrt{6}}{5}$: Uses $a$ instead of $2a$ in the denominator.

The denominator of the quadratic formula is $2a$; to reduce, divide every term of the numerator and the denominator by the same factor.

Answer: D

1. 1. Add 3 to both sides: $x^2+10x=3$.
2. 2. Half of 10 is 5, and $5^2=25$; add 25 to both sides: $x^2+10x+25=28$.
3. 3. Factor: $(x+5{)}^2=28$.

$(x-5{)}^2=22$: Wrong sign of $p$ and wrong constant: $+10x$ gives $(x+5{)}^2$, and 3 must be added to 25.

$(x-5{)}^2=28$: Wrong sign of $p$: $(x-5{)}^2=x^2-10x+25$.

$(x+5{)}^2=22$: Subtracts 3 from 25 instead of adding: after moving $-3$, the right side is $3+25$.

$(x+5{)}^2=28$: Correct. $x^2+10x=3$; add ${\left(\frac{10}{2}\right)}^2=25$ to both sides: $(x+5{)}^2=28$.

Move the constant first, then add ${\left(\frac{b}{2}\right)}^2$ to both sides; the sign inside the square matches the sign of $b$.

Answer: A

1. 1. $x^2-5x=0$.
2. 2. $x(x-5)=0$.
3. 3. $x=0$ or $x=5$.

Her method loses the solution $x=0$, because dividing by $x$ is not valid when $x=0$.: Correct. $x^2-5x=0$ gives $x(x-5)=0$, so $x=0$ or $x=5$; dividing by $x$ assumes $x\ne 0$ and discards 0.

Her method is correct, and 5 is the only solution.: $0^2=5(0)$, so 0 is also a solution.

Her method loses the solution $x=-5$, because a square has two square roots.: $(-5{)}^2=25$ but $5(-5)=-25$, so $-5$ is not a solution.

Her method is correct, because both sides were divided by the same expression.: This is true only for a nonzero divisor; $x$ can be 0, so the method loses a solution.

Do not divide both sides by a variable expression; move all terms to one side and factor instead.

Answer: C

1. 1. $P(x)=(-2x^2+60x)-(2x^2+12x+100)$.
2. 2. Distribute the minus sign: $-2x^2+60x-2x^2-12x-100$.
3. 3. Combine: $-4x^2+48x-100$.

$72x+100$: Adds $R(x)$ and $C(x)$ instead of subtracting.

$-4x^2+72x+100$: Subtracts only $2x^2$; the $12x$ and $100$ of the cost must also be subtracted.

$-4x^2+48x-100$: Correct. $-2x^2+60x-2x^2-12x-100=-4x^2+48x-100$.

$4x^2-48x+100$: Computes $C(x)-R(x)$, the opposite of profit.

Profit is revenue minus cost; put the whole cost polynomial in parentheses so every term is subtracted.

Answer: B

1. 1. GCF: $2x^4-32=2(x^4-16)$.
2. 2. $x^4-16=(x^2{)}^2-4^2=(x^2+4)(x^2-4)$.
3. 3. $x^2-4=(x+2)(x-2)$.
4. 4. $2(x^2+4)(x+2)(x-2)$.

$2(x^2+4)(x^2-4)$: Equivalent, but $x^2-4$ is still a difference of squares, so it is not factored completely.

$2(x^2+4)(x+2)(x-2)$: Correct. $2(x^4-16)=2(x^2+4)(x^2-4)=2(x^2+4)(x+2)(x-2)$; $x^2+4$ cannot be factored further.

$2(x+2{)}^2(x-2{)}^2$: Treats $x^2+4$ as if it were $x^2-4$; this expands to $2x^4-16x^2+32$.

$(2x^2+8)(x+2)(x-2)$: Equivalent, but the GCF 2 is left inside $2x^2+8$, so it is not factored completely.

Factor completely means no factor can be factored again: take out the GCF and keep applying difference of squares; a sum of squares does not factor.

Answer: A

1. Differences in $y$: $9,36,144$ — not constant, so not linear.
2. Ratios: $12/3=4$, $48/12=4$, $192/48=4$ — constant over each interval of 2.
3. So $y=3(2{)}^x$: the factor per unit of $x$ is 2, and per 2 units it is $2^2=4$.

It is exponential, because $y$ is multiplied by 4 each time $x$ increases by 2.: Correct. $12÷3=48÷12=192÷48=4$ over each interval of 2, so the function grows by equal factors over equal intervals.

It is linear, because $x$ increases by 2 each time.: Equal steps in $x$ are needed to test either model; what matters is how $y$ changes, and the differences 9, 36, 144 are not equal.

It is exponential, because $y$ is multiplied by 4 each time $x$ increases by 1.: Right model, wrong interval: the factor 4 applies to each increase of 2 in $x$, so the factor per 1 unit is 2.

It is linear, because the differences in $y$, 9, 36 and 144, change by a constant factor.: A linear function has equal differences; differences that are multiplied by the same factor each time are a sign of exponential growth, not linear growth.

Check the change in y over equal x-intervals: a constant difference means linear, a constant ratio means exponential. Note the interval length when stating the factor.

Answer: B

1. $[0,2]$: $160÷2=80$.
2. $[2,6]$: $600÷4=150$.
3. $[5,10]$: $480÷5=96$.
4. $[0,10]$: $1000÷10=100$. Greatest: $2\le w\le 6$.

$0\le w\le 2$: $\frac{260-100}{2}=80$ downloads per week, the least of the four.

$2\le w\le 6$: Correct. $\frac{860-260}{6-2}=\frac{600}{4}=150$ downloads per week, the greatest.

$5\le w\le 10$: $\frac{1100-620}{5}=96$ downloads per week.

$0\le w\le 10$: Has the greatest total change (1000) but over 10 weeks, so the rate is only 100 downloads per week.

The greatest average rate of change is the greatest change per unit of input, not the greatest total change; divide each change by its interval length.

Answer: A

1. $h(0)=112$, so the graph starts at $(0,112)$.
2. Vertex: $t=\frac{-96}{2(-16)}=3$, $h(3)=256$.
3. Lands when $h(t)=0$: $t=7$ (see the next item).
4. Only Graph A shows all three features with $0\le t\le 7$.

Graph A: Correct. Starts at $(0,112)$, the cliff height; peaks at $(3,256)$; and ends at $(7,0)$ when the rocket lands.

Graph B: Starts at $(0,0)$: this is $-16t^2+96t$, which leaves out the 112-foot starting height.

Graph C: The parabola is right, but it starts at $t=-1$; time before the launch is not in the domain, which is $0\le t\le 7$.

Graph D: Starts at 96 and peaks near $t=3.5$: this swaps the 96 and the 112, using $-16t^2+112t+96$.

For a projectile model h(t) = −16t² + bt + c, the graph starts at (0, c), peaks at t = −b/(2a), and ends at the positive zero; time cannot be negative.

Answer: A

1. List the pairs: $(-2,4)$, $(0,4)$, $(3,7)$, $(5,7)$.
2. No input appears twice, so each input has exactly one output: the relation is a function.

It is a function, because each input is paired with exactly one output.: Correct. Each of $-2$, 0, 3 and 5 has exactly one arrow. Two inputs sharing an output is allowed in a function.

It is not a function, because two inputs are paired with the same output.: Confuses the rule: a function cannot send one input to two outputs, but different inputs may share an output.

It is a function, because each output is paired with exactly one input.: Right verdict, false reason: the output 4 is paired with both $-2$ and 0. The definition is about inputs, not outputs.

It is not a function, because no single equation gives each output from its input.: A function does not need a formula; any relation in which each input has exactly one output is a function, and this mapping pairs each of $-2$, 0, 3 and 5 with one output.

A relation is a function when every input (element of the domain) is paired with exactly one output; outputs may repeat.

Answer: 800 square feet

1. Fencing: $2x+(L-20)=60$, where $L$ is the side opposite the barn, so $L=80-2x$.
2. Area: $A(x)=x(80-2x)=-2x^2+80x$, so $a=-2$, $b=80$.
3. Axis of symmetry: $x=\frac{-b}{2a}=\frac{-80}{2(-2)}=20$.
4. $A(20)=20(80-40)=20(40)=800$. The side opposite the barn is 40 feet, which is longer than the 20-foot hedge, so the design works.
5. The maximum area is 800 square feet.

Write the unknown side in terms of x from the fencing constraint (the hedge adds 20 feet of length for free), then find the maximum at x = −b/(2a). Answering 20 gives the side length, and using 60 − 2x for the long side ignores the hedge (maximum 450). On the Regents, a correct answer with no work shown earns only 1 of 2 credits.

Answer: C

1. Condition: attended the planetarium. Column total: 114.
2. Children who attended: 66.
3. $\frac{66}{114}\approx 0.5789\approx 57.9\%$.

$30.0\%$: $\frac{66}{220}$: the joint relative frequency (children who attended, out of all visitors), not the percent of attendees.

$45.5\%$: $\frac{100}{220}$: the percent of all visitors who were children, ignoring attendance.

$57.9\%$: Correct. The condition "of the visitors who attended" makes the column total, 114, the denominator: $\frac{66}{114}\approx 0.579=57.9\%$.

$66.0\%$: $\frac{66}{100}$: the percent of children who attended, which uses the row total instead of the column total.

For a conditional relative frequency, the group named after "of the" (or "given") supplies the denominator.